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Homework Help: Find the General Anti-derivative (Calculus I)

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data
    given f(x) = [x^3+5sqrt(x)]/x^2, find the anti-derivative

    2. Relevant equations

    3. The attempt at a solution

    Hi I have attempted to solve this by re-writing the equation as a sum of two fractions:

    x^3/x^2 + 5sqrt(x)/x^2, simplying gives = x + 5/x^3/2

    I then apply the principle anti-differentiation rules, and I come out with:

    x^2/2 + 5x^-1/2 = x^2/2 + 10/sqrt(x) + C

    This answer is coming back in WebWork as wrong. I also checked my answer on wolframalpha, and got the same result.

    What am I doing wrong? (If anything)
  2. jcsd
  3. Apr 29, 2012 #2


    User Avatar
    Gold Member

    Try re-writing it: [tex]x + \frac{5}{x^{3/2}}=x+5x^{-3/2}[/tex]Then integrate.
  4. Apr 29, 2012 #3
    I am not quite at intergration yet, this is the last section of my Calculus I course, Calculus II is intergration. So for now our anti-d's are pretty simple and just follow some basic rules.
  5. Apr 29, 2012 #4
    Ok so
    x + 5x-3/2 following the rule: x(n+1)/(n+1) gives

    x1+1 / (1+1) + 5x-3/2 + 2/2 / -(1/2) =

    x2/2 + 5x-1/2/ -(1/2 ) = -10x-1/2 or finally

    x2 / 2 + 10 / sqrt(x)

    I am still doing it wrong?
  6. Apr 29, 2012 #5
    ok yes I am doing it wrong, its - 10 / sqrt(x).

    Thank you very much.

    Story of my life, I always miss a sign.
  7. Apr 29, 2012 #6


    Staff: Mentor

    Integration is finding the antiderivative.

    Note that there is no such word in English as intergration.
  8. Apr 29, 2012 #7
    Thank you Mark for the insight. And you are right, there is no such word as intergration.
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